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**Design of foundation for Fattouh building in Nablus**

Prepared by Israa Hanani Noor Sobuh Nibal Awwad Supervisor Dr: Sami Hijjawi

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**Outline 1- Introduction and site description.**

2- Load description and type of foundation. 3- Design of foundation. a. single footing b. mat foundation c. pile foundation

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**Introduction and site description**

The building foundation or sub structure is that part of a structure which is placed below the surface of the ground and which transmits the superstructure load to the underlying soil ultimately . It is the part of a structural system that supports and anchors the superstructure of a building . Building located Fattouh for Aljnied prison and consists of 7 floors. Soil type built by architecture is clay soil . And the allowable bearing capacity is 182 KN/m2. Based on soil testing, the origin was designed to foundations mat foundation and under it pile .

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**Load description The proposed building has 23 type**

of columns based on their dimension . The distribution of these columns and there corresponding load are shown in table. As shown from the table below, the columns service loads range from Tons . Col. no Dimension Load (KN) C1 20x60 2110 C4 50x50 2235 C17 30x160 2170 C18 20X100 1540 C19 20x175 2940 C23 20x120 1314

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Type of foundation Type Pile Mate Single

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**Design single footing For column 1 (as an example ) :**

The dimension of single footing according to applied service load and bearing capacity of soil as follow : For column 1 (as an example ) : Area of footing = Pall / Q all = 2110 / 182 =11.6 m2 (4.5m * 3m) and repeat it for all column . From the Fig. we see the overlap between the adjacent footing, so the single footing doesn't fit for this project.

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Mat Foundation Area of mat footing = m2

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**1) We calculate center of loads X p= 1353055.8/ 65563 = 20.64 m **

Calculate the center of loading that applied at columns (col. 1 as an example) : column Name Service KN))load Yi Xi P*Xi P*Yi C1 2110 11.65 39.9 84189 After that 1) We calculate center of loads X p= / = m Yp= / = m 2) Calculate the eccentricity in X and Y directions ex = YP – YG = – = m ey = XP – XG = – 21.3 = m

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**Total soil pressure = 4.53* 13.45 *113.13**

q= -(Q/A) – (My X/ Iy ) – (Mxy/Ix) Ix = bh3 / = m4 Iy = m Mx = Q ex = KN.m My = KN.m q = 125 < 182 KN/m2 ……. Ok We take internal stripe in y- direction with width 4.53m as shown in fig. above Total soil pressure = 4.53* *113.13 = KN ∑ col loads = = 6818 KN

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Shea force diagram

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**B.M.D. Mu = 3615.5 KN.m From the figure of shear and moment we do the**

checks and compare the result to the max. value

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**Calculate the thickness of mat footing:-**

Width of strip 4.53m Vu = KN Assume D =800 mm h =900 mm Wide beam shear check :- Ø vc = (0.75 / 6) (28 ) 0.5 *4530 * 800 / 1000 = KN > Vu ok Mu = KN.m ρ = > ρ min ok As = ρ bd = * 4530 *800 = mm 2 > As min 1Ø 25 / 200 mm As min = bh = mm2

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**Check Punching :- Check punching for C 8 :- Pu = 1.4 * 3230 = 4522 KN**

ØVc = ( Ø /3 ) (fc )0.5 bod = ( 0.75 / 3) (28)0.5(2* * 1200 ) (800 ) /1000 = KN > Pu For columns 11 and 10 we may use drop panel or increase the amount of steel to make it ok for punching

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Pile Footing A pile is a relatively long columnar construction element made of wood, or reinforced concrete, or metal, or a combination thereof. It is embedded into soil to receive and transmit vertical and transmit vertical and inclined loads into the soil or to rock below the ground surface

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**Design Pile Foundation For Column 11**

Use All-Pile program Total Ultimate Capacity (Down)= kN, Total Ultimate Capacity (Up)= kN Total Allowable Capacity (Down)= kN, Total Allowable Capacity (Up)= kN

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**Design Pile Foundation For Column 11**

where it's service load equal KN/m2 Number of Pile Find the min. length for pile from the equation and satisfied the min. Ep RI / Su = So from table min. length = 11* D = 8.8 m Assume allowable capacity of a 0.8-m-diameter and m-long pile Total Allowable Capacity (Down)= kN F=( P/N) +( Mxy/Σyi2) + (Myx/Σxi2) ≤ qall F= (5788)/N ≤ → N=8→ Use 8 Piles

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** The PLAN show the pile distribution and the spacing**

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**Thickness of cap: Pu = 1. 2 DL + 1. 6LL = 7813. 8 KN Ru =Pu/N = 7813**

Thickness of cap: Pu = 1.2 DL + 1.6LL = KN Ru =Pu/N = /8 = KN Wide beam shear : Vu =976.72*2 = KN ϕVc = (1/6) (√fc) bd /1000 = 0.75(1/6)(√28)(3500)d/1000= → d = 850 mm → h =1050 mm

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Punching shear : For column: Vu= KN ϕVc= 0.75(1/3)(√28)*800*(1350*2+1350*2)/1000= 6071KN > KN → NOT OK SO we increase d= 1000 mm , h=1200 mm ϕVc = 7937 > KN → OK

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**Reinforcement Long direction**

Mu = Pu L / 4 = 1172 KN.m , d= 1000 mm, ρ= * 10-3 As = *10-3* 3500 * 1000 = 3122 mm2 As min = bh = * 3500 * 1200 = 7560 mm For bottom steel > As → Use As min → use 15 ϕ 25 For Top Steel ->> 7560/2 = 3780 mm2 use 12 ϕ 20 Short direction = *3500*1200 = 7560 mm2 >As Use As min → use 15 ϕ 25

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**From Table at Dia. 800 mm Mcapacity = 901.5 KN.m As = 25.1 (10 φ 18)**

The figure show the distribution of the steel (a) Show the distribution steel for cap from analysis and (b) Show the distribution of the steel for Pile from classification φ

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